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Minchanka [31]
2 years ago
11

What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters? Express your answer to the nearest w

hole number. 156.25 meters per second
Physics
2 answers:
qwelly [4]2 years ago
7 0

Answer:

The speed of the wave is 156.25 meter per second .

Explanation:

Use the formula for the relation between the speed of a wave and frequency.

v=f \lambda

Where f is frequency and \lambda is wavelength .

f = 125 hz

\lambda=1.25\ meters

Putting all the values in the formula

v=125\times 1.25

v=156.25\ meter\ per second

Therefore the speed of the wave is 156.25 meter per second .

shtirl [24]2 years ago
3 0

156 is the answer. so 156.25 is almost the same thing, you just round. It's not hard. Thank you!!!

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2 years ago
Section 2.2
Feliz [49]

Answer:

Speed changes at the rate of 24 m/s for each second over time.

Explanation:

We are told the object's acceleration is equal to 24 m/s²

Now we know that acceleration can also be defined as the rate of change of speed with time. Also speed has a unit known as m/s.

Thus, we can rephrase the acceleration in this question to mean;

Speed changes at the rate of 24 m/s for every second with time.

6 0
3 years ago
The gage pressure in a liquid at a depth of 3 m is read to be 39 kPa. Determine the gage pressure in the same liquid at a depth
ioda

Answer: 117 kPa

Explanation:

For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa

For the liquid at depth 9m, the gauge pressure is equal to= P₂

Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.

So, For finding gauge pressure we have formula P= ρ * g * h

Also gravity also remains same for both liquids

So taking ratio of their respective pressures we have

\frac{P_{1} }{\\P_2}= \frac{density * g * h_1}{density * g * h_2}

So \frac{39}{P_2}= \frac{3}{9}

Or P₂= 39 * 3 = 117 kPa

5 0
3 years ago
2.)
Oduvanchick [21]

Answer:

-22.7 m/s^2

Explanation:

This is a uniformly accelerated motion, so we can determine the deceleration of the car by using a suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the car in this problem,

u = 27.8 m/s

v = 0

s = 17 m

Solving for a, we find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{0-27.8^2}{2(17)}=-22.7 m/s^2

4 0
3 years ago
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